Why is the sum of squares of 1~k =1/6 * k (k+1) * (2k+1)?

Prove by recursive induction.

1/6 * k(k+ 1)*(2k+ 1)= 200 * N

k(k+ 1)*(2k+ 1)= 1200 * N = 4 * 3 * 4 * 25 * N

Multiply both sides by 4.

2k * (2k+1) (2k+2) = 4 * 4 * 3 * 25 * n = the product of three consecutive numbers, in which there must be a multiple of 3.

(2K)(2K+2) has a multiple of 32;

2K+ 1 is an odd factor, a multiple of 32 or a multiple of 25.

So the multiple of 32 is close to the multiple of 25: if there is 224, then k= 1 12 can be used.

So the minimum k is 1 12.